I'm Stumped on this Trig Question?

Question

While hiking on a level path, Otis determines that the angle of elevation to the top of Long's Peak is 30 degrees. Moving 1000 feet closer to the mountain, Otis determines the angle of elevation to be 35 degrees. How much higher is the top of Longs Peak than Otis's Elevation?

Answer 1

At the start, Otis is twice the elevation away from the cliff's base.  If he moves 1000 ft. closer and is now 5 degrees "closer," we can assume that 200 ft = 1 degree growth, from 30 degrees.

So from 30 degrees, he's 6000 ft. out (200 * 30).  The elevation must be 1/2 that, as it would be left being 60 degrees, assuming the peak and the level ground are directly in line (and thusly form a 90-degree angle).

Cliff elevation equals one-half Otis' distance to the cliff base.

1/2 of 6000 is 3000.

Long's Peak is 3000 ft above Otis' elevation.

Please see the illustration for more detailed information.

http://i21.photobucket.com/albums/b275/A...

Answer 2

Let h = height of the mountain

=> h(cot30° - cot35°) = 1000

=> h

= 1000/(cot30° - cot35°)

= 303.9 ft.

Answer 3

A more simplified way of looking at it than HopelesRomantc91's view (though the same idea).

Where 1/2 the total distance equals 15 degrees, 1/3 of that half (1/6th) would be 5 degrees.  If 1/6th = 5 degrees = 1000ft, and elevation = 1/2 total distance, elevation equals 3 times 1000ft.

Elevation equals 3000 ft.

Just a little geometry and Pythagoras.  Nothing too complicated about that.  I haven't even taken math classes past Alg. 2, and I didn't even need a pen and paper, none the less scientific/graphing calculator, to solve that problem.  Have fun!

Answer 4

30            35

          x

1000         y

30y = 1000x35

30y   35000

      =

30     30

y = 1166.66....

so Longs Peak is that much higher 1166.66....  

Answer 5

I do not understand the question.  But let us try to determine the height of Long's Peak:

Let y be the height of Long's Peak

Let x be the distance from Otis to the base of the mountain where the 30 deg measurement was taken.  So then the distance from the 35 degree measurement to the base is x - 1000 feet.

sin 30 = y/x

sin 35 = y/(x-1000)

There are two eqns in two unknowns and

y = x sin 30

y = (x-1000) sin 35

x sin 30 = (x-1000) sin 35 = x sin 35 - 1000 sin 35

x (sin 30 - sin 35) = - 1000 sin 35

x = -1000 sin 35/(sin 30 - sin 35)

This is the number value of x then y, the height is

y = x sin 30

I will leave it to you to put in the numbers